EQUILIBRIA WITH SALTS
Dissolving Rate
Crystallizing Rate
SATURATED – A solution where the solid and dissolved forms of the solute
are in equilibrium, and contains as much dissolved solute as possible
SOLUBILITY – The maximum amount of solute that will dissolve in a
specific amount of solvent
MOLAR SOLUBILITY – The solubility of a solute, measured in mol/L
2D-1 (of 11)
SOLUBILITY PRODUCT CONSTANT (Ksp) – The equilibrium constant for
the reaction of a solid salt dissolving in water
silver chloride
AgCl (s) ⇆ Ag+ (aq) + Cl- (aq)
Ksp = [Ag+][Cl-]
calcium phosphate
Ca3(PO4)2 (s) ⇆ 3Ca2+ (aq) + 2PO43- (aq)
Ksp = [Ca2+]3[PO43-]2
2D-2 (of 11)
CALCULATING THE SOLUBILITY PRODUCT CONSTANT (Ksp)
The solubility of calcium fluoride is 0.017 g/L. Calculate its Ksp.
CaF2 (s)
⇆
Initial M’s
Change in M’s
Equilibrium M’s
0
+x
x
(- x)
Ksp = [Ca2+][F-]2
Ca2+ (aq)
= (x)(2x)2 = 4x3
x = molar solubility of CaF2
0.017 g CaF2 x
mol CaF2
________________
________________
L
78.08 g CaF2
Ksp = 4(2.18 x 10-4 M)3
2D-3 (of 11)
=
= 2.18 x 10-4 M CaF2
4.1 x 10-11 M3
+
2F- (aq)
0
+ 2x
2x
The solubility of iron (III) hydroxide is 0.013 g/L. Calculate its Ksp. Molar
mass of iron (III) hydroxide is 106.88 g/mol.
Fe(OH)3 (s)
Initial M’s
Change in M’s
Equilibrium M’s
⇆
Fe3+ (aq)
0
+x
x
(- x)
+
3OH- (aq)
0
+ 3x
3x
Ksp = [Fe3+][OH-]3 = (x)(3x)3 = 27x4
x = molar solubility of Fe(OH)3
0.013 g Fe(OH)3 x
____________________
L
mol Fe(OH)3
______________________
106.88 g Fe(OH)3
Ksp = 27(1.22 x 10-4 M)4
2D-4 (of 11)
= 1.22 x 10-4 M Fe(OH)3
=
5.9 x 10-15 M4
PREDICTING RELATIVE SOLUBILITIES
Which is more soluble, BaF2 (Ksp = 1.4 x 10-7) or MgF2 (Ksp = 6.4 x 10-9)?
The bigger molar solubility, the more soluble the salt
The bigger the Ksp, the bigger the molar solubility  BaF2 more soluble
… provided the number of ions per formula unit are identical
MF2 (s)
Initial M’s
Change in M’s
⇆
(- x)
Equilibrium M’s
Ksp = [M2+][F-]2 = (x)(2x)2 = 4x3
3
Ksp
____
4
2D-5 (of 11)
= x
= molar solubility of MF2
M2+ (aq)
0
+x
x
+
2F- (aq)
0
+ 2x
2x
Which is more soluble, Ag2S (Ksp = 1.6 x 10-49) or CuS (Ksp = 8.5 x 10-45)?
The number of ions per formula unit are not identical
Ag2S (s) ⇆ 2Ag+ (aq) + S2- (aq)
I
C
E
(- x)
0
+ 2x
2x
0
+x
x
Ksp = [Ag+]2[S2-]
CuS (s) ⇆ Cu2+ (aq) + S2- (aq)
(- y)
0
+y
y
0
+y
y
Ksp = [Cu2+][S2-]
1.6 x 10-49 = 4x3
8.5 x 10-45 = y2
3.4 x 10-17 = x = molar solubility
9.2 x 10-23 = y = molar solubility
Bigger molar solubility, so Ag2S is more soluble
2D-6 (of 11)
CALCULATING THE SOLUBILITY OF A COMPOUND
Find the molar solubility of calcium sulfate if its Ksp = 2.4 x 10-5.
CaSO4 (s)
⇆
Initial M’s
Change in M’s
Equilibrium M’s
(- x)
Ca2+ (aq)
+
0
+x
x
SO42- (aq)
0
+x
x
Ksp = [Ca2+][SO42-]
2.4 x 10-5 = x2
x = 4.90 x 10-3 M = molar solubility of CaSO4
Find the solubility of calcium sulfate in g/L.
4.90 x 10-3 mol CaSO4 x 136.15 g CaSO4
____________________________
L
2D-7 (of 11)
____________________
mol CaSO4
= 0.67 g/L
Find the molar solubility of lead (II) iodide if its Ksp = 8.3 x 10-9.
PbI2 (s)
Initial M’s
Change in M’s
Equilibrium M’s
(- x)
⇆
Pb2+ (aq)
0
+x
x
Ksp = [Pb2+][I-]2
8.3 x 10-9 = (x)(2x)2 = 4x3
x = 1.28 x 10-3 M = molar solubility of PbI2
2D-8 (of 11)
+
2I- (aq)
0
+ 2x
2x
Find the molar solubility of lead (II) iodide in a 0.10 M sodium iodide
solution.
The NaI contains a common ion in the PbI2 equilibrium reaction
PbI2 (s)
Initial M’s
Change in M’s
Equilibrium M’s
⇆
(- x)
Pb2+ (aq)
0
+x
x
+
2I- (aq)
0.10
+ 2x
0.10 + 2x
Ksp = [Pb2+][I-]2
8.3 x 10-9 = (x)(0.10 + 2x)2 = (x)(0.10)2
x = 8.3 x 10-7 M
= molar solubility
The I- shifts the equilibrium to the left, making PbI2 less soluble in this
solution than in pure water (8.3 x 10-7 M < 1.3 x 10-3 M)
2D-9 (of 11)
PREDICTING PRECIPITATION
If a solution is 0.10 M in sodium hydroxide and 0.15 M in barium nitrate,
determine if a precipitate of barium hydroxide will form.
NaOH completely dissociates
 0.10 M Na+ and 0.10 M OH-
Ba(NO3)2 completely dissociates  0.15 M Ba2+ and 0.30 M NO3Ba(OH)2 (s)
Initial M’s
Change in M’s
Equilibrium M’s
Ksp = [Ba2+][OH-]2
Q
Q < Ksp
?
⇆
Ba2+ (aq)
+
0.15
?
2OH- (aq)
0.10
?
Look up the Ksp: 5.0 x 10-3
= [Ba2+][OH-]2 = (0.15 M)(0.10 M)2 = 1.5 x 10-3
 forward reaction spontaneous  no precipitate forms
2D-10 (of 11)
If a solution is 0.010 M in nickel (II) nitrate and 0.025 M in sodium carbonate,
determine if a precipitate of nickel (II) carbonate (Ksp = 1.4 x 10-7) will form.
Ni(NO3)2 completely dissociates  0.010 M Ni2+ and 0.020 M NO3Na2CO3 completely dissociates
 0.050 M Na+ and 0.025 M CO32-
NiCO3 (s)
Initial M’s
Change in M’s
Equilibrium M’s
?
⇆
Ni2+ (aq)
0.010
?
+
CO32- (aq)
0.025
?
Ksp = [Ni2+][CO32-]
Q
Q > Ksp
= [Ni2+][CO32-] = (0.010 M)(0.025 M) = 2.5 x 10-4
 reverse reaction spontaneous  a precipitate forms
2D-11 (of 11)
PREDICTING ONE EQUILIBRIUM ION CONCENTRATION
Calculate the concentration of fluoride ions needed to produce a precipitate
in a 0.100 M barium nitrate solution. The Ksp of barium fluoride is 1.4 x 10-7.
Equilibrium is when the precipitate just starts to form  want the [F-]eq
Ba(NO3)2 completely dissociates  0.100 M Ba2+ and 0.200 M NO3BaF2 (s)
⇆
Ba2+ (aq)
+
2F- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
Ksp = [Ba2+] [F-] 2
0.100
1.4 x 10-7
1.3 x 10-3 M
2E-1 (of 11)
=
x
(0.100) x2
= x = [F-]eq
Crystals of NaOH are added to a solution that is 0.100 M Sr(NO3)2 until the
OH- concentration is 0.200 M. Calculate the percentage of Sr2+ ions left in
the solution at this point. The Ksp of Sr(OH)2 is 3.2 x 10-4.
The answer is not 0% because the reaction does not go to completion,
it reaches an equilibrium  solve for [Sr2+]eq
Sr(OH)2 (s)
Initial M’s
Change in M’s
Equilibrium M’s
⇆
Sr2+ (aq)
+
x
2OH- (aq)
0.200
Ksp = [Sr2+][OH-] 2
3.2 x 10-4
=
x(0.200)2
0.0080 M x 100 = 8.0%
0.0080 M
=
x
0.100 M
2E-2 (of 11)
___________
PREDICTIONS INVOLVING TWO PRECIPITATES
Ksp for AgOH is 2.0 x 10-8 and Ksp for Mn(OH)2 is 2.0 x 10-13.
(a) If hydroxide ions are added to a solution 0.100 M Ag+ and 0.200 M
Mn2+, what will precipitate first?
Solve for [OH-]eq needed to precipitate each
AgOH (s) ⇆ Ag+ (aq) + OH- (aq)
Mn(OH)2 (s) ⇆ Mn2+ (aq) + 2OH- (aq)
0.100
0.200
I
C
E
x
Ksp = [Ag+][OH-]
Ksp = [Mn2+][OH-]2
2.0 x 10-8
= (0.100)x
2.0 x 10-13 = (0.200)(x)2
2.0 x 10-7
= x = [OH-]eq to ppt
1.0 x 10-6
= x = [OH-]eq to ppt
AgOH will precipitate first because it requires a smaller [OH-]eq
2E-3 (of 11)
x
PREDICTIONS INVOLVING TWO PRECIPITATES
Ksp for AgOH is 2.0 x 10-8 and Ksp for Mn(OH)2 is 2.0 x 10-13.
(b) What will be the concentration of Ag+ when the Mn(OH)2 starts to
precipitate?
[OH-]eq needed to precipitate Mn(OH)2 was 1.0 x 10-6 M
AgOH (s)
Initial M’s
Change in M’s
Equilibrium M’s
=
0.020 M =
2E-4 (of 11)
x(1.0 x 10-6)
x
Ag+ (aq)
x
Ksp = [Ag+][OH-]
2.0 x 10-8
⇆
=
[Ag+]
+
OH- (aq)
1.0 x 10-6
EFFECT OF pH ON PRECIPITATION
If a solution is saturated with H2S (0.10 M) and the pH adjusted to 1.00 with
HCl, determine the molar solubility of FeS.
Whenever a solution has its “pH adjusted”, this means an external source
of acid or base has been added
In this case, adjusting the pH controls the [S2-]
Ksp for FeS = 3.7 x 10-19
Ka1 for H2S = 1.0 x 10-7
Ka2 for H2S = 1.0 x 10-15
Use the Ka1,2 to determine [S2-]
Ka1,2 = Ka1Ka2
2E-5 (of 11)
= (1.0 x 10-7)(1.0 x 10-15)
= 1.0 x 10-22
EFFECT OF pH ON PRECIPITATION
If a solution is saturated with H2S (0.10 M) and the pH adjusted to 1.00 with
HCl, determine the molar solubility of FeS.
[H3O+]eq = antilog (-1.00)
= 0.10 M
H2S (aq)
Initial M’s
Change in M’s
Equilibrium M’s
Ka1,2 = [H3O+]2[S2-]
______________
[H2S]
+
2H2O (l)
⇆
2H3O+ (aq)
0.10
0.10
1.0 x 10-22 =
(0.10)2x
____________
(0.10)
1.0 x 10-21 M = x = [S2-]eq
2E-6 (of 11)
+
S2- (aq)
x
EFFECT OF pH ON PRECIPITATION
If a solution is saturated with H2S (0.10 M) and the pH adjusted to 1.00 with
HCl, determine the molar solubility of FeS.
FeS (s)
⇆
Initial M’s
Change in M’s
Equilibrium M’s
Ksp = [Fe2+][S2-]
3.7 x 10-19 = (x)(1.0 x 10-21 M)
370 M = x = molar solubility of FeS
 FeS is very soluble at a pH = 1.00
2E-7 (of 11)
Fe2+ (aq)
x
+
S2- (aq)
1.0 x 10-21
If a solution is saturated with H2S (0.10 M) and the pH adjusted to 10.00 with
NaOH, determine the molar solubility of FeS.
[H3O+]eq = antilog (-10.00)
= 1.0 x 10-10 M
H2S (aq)
Initial M’s
Change in M’s
Equilibrium M’s
Ka1,2 = [H3O+]2[S2-]
______________
[H2S]
0.10
+
2H2O (l)
⇆
2H3O+ (aq)
1.0 x 10-10
1.0 x 10-22 = (1.0 x 10-10 )2x
_________________
(0.10)
1.0 x 10-3 M = x = [S2-]eq
2E-8 (of 11)
+
S2- (aq)
x
If a solution is saturated with H2S (0.10 M) and the pH adjusted to 10.00 with
NaOH, determine the molar solubility of FeS.
FeS (s)
⇆
Initial M’s
Change in M’s
Equilibrium M’s
Fe2+ (aq)
x
Ksp = [Fe2+][S2-]
3.7 x 10-19 = (x)(1.0 x 10-3 M)
3.7 x 10-16 M = x = molar solubility of FeS
 FeS is very insoluble at a pH = 10.00
2E-9 (of 11)
+
S2- (aq)
1.0 x 10-3
Calculate the pH that FeS precipitates in a solution 0.10 M Fe2+ and
saturated with H2S (0.10 M).
Use the Ksp to determine the [S2-]eq needed to precipitate FeS
FeS (s)
⇆
Fe2+ (aq)
+
S2- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
Ksp = [Fe2+][S2-]
0.10
3.7 x 10-19 = (0.10)x
∴ [S2-]eq = 3.7 x 10-18 M to start to precipitate FeS
Use the Ka1,2 to determine the pH
2E-10 (of 11)
x
3.7 x 10-18 M = x
Calculate the pH that FeS precipitates in a solution 0.10 M Fe2+ and
saturated with H2S (0.10 M).
H2S (aq) + 2H2O (l) ⇆ 2H3O+ (aq) + S2- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
Ka1,2 = [H3O+]2[S2-]
______________
0.10
3.7 x 10-18
x
1.0 x 10-22 = (x)2(3.7 x 10-18)
__________________
[H2S]
(0.10)
1.64 x 10-3 = x
pH = -log (1.64 x 10-3) = 2.78
2E-11 (of 11)